[leetcode] Number of 1 Bits

最新推荐文章于 2019-02-27 11:14:59 发布

Cabinathor

最新推荐文章于 2019-02-27 11:14:59 发布

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分类专栏： Leetcode

本文链接：https://blog.youkuaiyun.com/lmxmimihuhu/article/details/46128027

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本文介绍了一种计算32位无符号整数中1的位数的方法，即所谓的汉明重量。通过两种不同的C++实现方式，一种是简单的逐位检查，另一种则是利用位操作技巧大幅提高效率。

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From : https://leetcode.com/problems/number-of-1-bits/

Write a function that takes an unsigned integer and returns the number of ’1' bits it has (also known as the Hamming weight).

For example, the 32-bit integer ’11' has binary representation 00000000000000000000000000001011, so the function should return 3.

class Solution {
public:
    int hammingWeight(uint32_t n) {
        int count = 0;
        while(n) {
            count += n&1;
            n >>= 1;
        }
        return count;
    }
};

Better Solution:

This is based on bit-manipulation, referenced by Dora' notes.

class Solution {
public:
    int hammingWeight(uint32_t n) {
        n = ((0xAAAAAAAA&n)>>1)+(0x55555555&n);
        n = ((0xCCCCCCCC&n)>>2)+(0x33333333&n);
        n = ((0xF0F0F0F0&n)>>4)+(0x0F0F0F0F&n);
        n = ((0xFF00FF00&n)>>8)+(0x00FF00FF&n);
        n = ((0xFFFF0000&n)>>16)+(0x0000FFFF&n);
        return n;
    }
};

public class Solution {
    // you need to treat n as an unsigned value
    public int hammingWeight(int n) {
		n = ((n & 0xAAAAAAAA) >>> 1) + (n & 0x55555555);
		n = ((n & 0xCCCCCCCC) >>> 2) + (n & 0x33333333);
		n = ((n & 0xF0F0F0F0) >>> 4) + (n & 0x0F0F0F0F);
		n = ((n & 0xFF00FF00) >>> 8) + (n & 0x00FF00FF);
		n = ((n & 0xFFFF0000) >>> 16) + (n & 0x0000FFFF);
		return n;
    }
}